From the statement we can conclude that a class can be slept through $$$\textbf{if and only if}$$$:

1. The class itself is not an important class.
2. The class is not within $$$k$$$ classes after any important class.

Here's an $$$O(nk)$$$ approach:

We directly simulate the process: for each important class at position $$$i$$$, mark positions $$$i+1, i+2, \dots, \min(i+k, n)$$$ as “must stay awake”. After processing all important classes, count the classes that are neither important nor marked. This $$$O(nk)$$$ solution is acceptable under the constraints.

There's also an $$$O(n)$$$ solution:

For each unimportant class, we only need to focus on the nearest important class before it. Thus we maintain a variable $$$last$$$ storing the index of the most recent important class (or $$$-\infty$$$ if none seen yet), then for each $$$i$$$ from $$$1$$$ to $$$n$$$, if it's an important class we let $$$last = i$$$, otherwise if $$$last + k \lt i$$$, the class is able to be slept through and we add $$$1$$$ to the answer.

Consider a simple greedy algorithm: For each turn, choose the option that gives the higher immediate score. That is, if our current score is $$$k$$$, it becomes $$$\max(k - a_i, b_i - k)$$$ after this turn. Why doesn't the algorithm work?

Try to improve the algorithm in Hint 1.

The greedy approach in Hint 1 fails because it assumes that if we know the maximum possible score after turn $$$i$$$ (denoted by $$$max_i$$$), we can determine $$$max_{i + 1}$$$ from it. However, this assumption is wrong and here is a counterexample:

$$$a = [2, 0], b = [1, 0]$$$

Here $$$max_1 = 1$$$ and $$$max_2 = 2$$$. The greedy approach takes the blue card in turn $$$1$$$. But to reach $$$max_2$$$, we must choose the red card in turn $$$1$$$, which turns our score into $$$-2$$$, and then choose the blue card in turn $$$2$$$. So in this example, $$$max_1$$$ does not give $$$max_2$$$.

We see that $$$max_1$$$ does not help us find $$$max_2$$$ if we choose the blue card in the second turn. The reason is, if we want to maximize our score when choosing a blue card, we want our current score to be as small as possible, so we are looking for the minimum possible score. To maximize $$$b_i - k$$$ we should minimize $$$k$$$.

From this we can see that if we have both the maximum possible score $$$max_{i - 1}$$$ and the minimum possible score $$$min_{i - 1}$$$ after turn $$$i - 1$$$, we can determine $$$max_{i} = \max(max_{i - 1} - a_i, b_i - min_{i - 1})$$$, because these two expressions cover the best outcomes from the red and blue options respectively. Similarly, the new minimum $$$min_{i} = \min(min_{i - 1} - a_i, b_i - max_{i - 1})$$$ comes from either taking the red card from the previous minimum, or the blue card from the previous maximum. And from $$$max_{i}, min_{i}$$$ we can move forward to $$$max_{i + 1}, min_{i + 1}$$$ and so on.

Thus, we start with $$$max_0 = min_0 = 0$$$ and calculate each $$$max_i, min_i$$$ throughout the process using the formulas above. The answer is $$$max_n$$$. This runs in $$$O(n)$$$ time complexity.

Consider the cases $$$n = 1$$$ and $$$n = 2$$$. What do you observe?

Try to determine the "first" element that must belong to $$$B$$$.

Assume the minimum element in $$$A$$$ is $$$x$$$. We now prove that $$$x$$$ must be included in $$$B$$$.

There are no elements smaller than $$$x$$$, which means $$$x$$$ has no proper divisor (other than $$$x$$$ itself) inside $$$A$$$. If we do not put $$$x$$$ into $$$B$$$, then the second rule would be violated.

Thus we can determine the first element of $$$B$$$, and then check whether all its multiples $$$\le k$$$ are contained in $$$A$$$. We mark these multiples as "finished".

Now think greedily. We never need to process elements that are already marked "finished". Each time, we just find the smallest unfinished element in $$$A$$$, check it, and add it to $$$B$$$ if possible.

A loose upper bound on the running time is $$$O\bigl(n d(n) \log n\bigr)$$$, where $$$d(n)$$$ is the divisor function. However, this bound is rarely attained in practice: many operations are skipped thanks to the "finished" marks, so the algorithm runs much faster than this worst-case estimate.

What is the answer when $$$k \geq 32$$$?

We can use dynamic programming to solve it.